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Special case when solving D.E.'s

  1. Feb 28, 2009 #1
    Why can't there be common terms on both complementary function and particular integral when solving differential equations?

    For instance,

    dy/dx + 3y = exp(-x) + exp(-3x)

    y(CF) = Aexp(−3x)

    y(PI) = Cexp(−x) + Dxexp(−3x)

    The term Dexp(-3x) in the P.I. has to be multiplied by x to be linearly independent of Aexp(-3x) in the C.F.. Why? What would happen if it wasn't?
     
  2. jcsd
  3. Mar 1, 2009 #2

    HallsofIvy

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    Try it and see! Take y(x)= C exp(-x)+ D exp(-3x) and put it into the equation. Try to solve for C and D to make it true.
     
  4. Mar 1, 2009 #3

    arildno

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    Note that due to the characteristics of the differentiation of a product, ALL products of the form Y(x)=f(x)e^(-3x) will experience some "internal cancellation".

    In your case, we will get on LHS:
    [tex]\frac{dY}{dx}+3Y=(f^{,}(x)e^{-3x}-3f(x)e^{-3x})+3f(x)e^{-3x}=f^{,}(x)e^{-3x}[/tex]
    Thus, since this must equal identically RHS, i.e, [tex]e^{-3x}[/tex], we get the subsidiary equation on f:
    [tex]f^{,}(x)=1[/tex], which is readily solved.
     
  5. Mar 1, 2009 #4
    D would vanish... That's why they have to be linearly independent? I can't think of it in practical situations.
     
  6. Mar 1, 2009 #5
    Where did you get this from?
     
  7. Mar 1, 2009 #6

    arildno

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    From defining Y(x) as f(x)e^(-3x), and applying the LHS operator on it!
     
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